The goal is to find the distribution of $T = \sum_{i=1}^n e^{-X_i}$ and also to compute $\textbf{E}(\log T)$ and $\textbf{V}(\log T)$.

Some thoughts:

I think I have found the distribution of $T$ by applying the transformation $Y = e^{-X}$. If I am not wrong, it is quite easy to see that
$Y \sim \textrm{Exponential}(e^{\theta})$. Therefore, $T = \sum_{i=1}^n Y_i \sim \textrm{Gamma} (n, 1/e^{\theta})$.

However, I am unable to find a reasonable way to compute $\textbf{E}(\log T)$ or $\textbf{V}(\log T)$. The formula for the expectation of a function of a random variable leads to a very complicated integral and the only alternative I can think of, which is yet another transformation, is even worse!

First of all, I would like to thank you for your answer, but there are some things that I don't understand. I don't know how to use $\log(x) = \frac{\mathrm{d}}{\mathrm{d}s} x^s ]_{s=0}$. In fact, I don't even know where it comes from. If it is an identity, I think I haven't seen it before, and if it is a change of variables, I don't know what to do. Again, sorry for the dumb questions, I've had a bad day and I can't seem to get anything!
–
Shiwen YaoSep 26 '11 at 19:38

Phew! I've been trying to reproduce the exercise on my own this morning, and I now I am stuck with the last identity $\mathbb{V}(\log T) = \left. \frac{ \mathrm{d}^2}{ \mathrm{d}s^2 } \mu^{-s} \mathbb{E}(x^s)\right\vert_{s=0}$. After computing $\mathbb{E}(\log T)$, I've just applied $\mathbb{V}(\log T) = \mathbb{E}(\log^2 T) - \mathbb{E}(\log T)^2$. I don't know if I got it right, but I couldn't simplify anything.
–
Shiwen YaoSep 27 '11 at 11:17